Orthogonal and Unitary transformatons; Rigid Motions
Definition: A matrix U such that U ∗ = U −1 is said to be unitary if its entries are complex,
and orthogonal if they are real.
Proposition:
1. U ∗ U = I ⇒ U U ∗ = I.
2. The rows of a unitary/orthogonal matrix written out in the standard basis form an
orthonormal basis for Cn (Rn ). So do the columns.
3. If H is self-adjoint, then exp(iH) is unitary.
Proposition: If U is unitary, then for all x, y ∈ V, ||U x − U y)|| = ||x − y||. That is, the
linear transformation x → U x preserves lengths.
Proof: We compute
||U x − U y)||2 = < U x − U y, U x − U y >
= < U (x − y), U (x − y) >
= < x − y, U ∗ U (x − y) >
= < x − y, x − y > = ||x − y||2
A unitary transformation also preserves the scalar product between two vectors:
< U x, U y > = < x, U ∗ U y >
= < x, y >
So if U is orthogonal, it preserves angles as well as distances, since
< U x, U y > < x, y >
= = cos(θ),
||U x|| ||U y|| ||x|| ||y||
where θ is the angle between the two vectors.
Definition: Let F : V → V be a function with the property ||F (x) − F (y)|| = ||x −
y||, ∀x, y ∈ V . Then F is said to be a rigid motion of V .
We already know that unitary transformations are rigid motions. They are also linear trans-
formations.
Definition: A transformation of the form Fa (x) = x−a, where a 6= 0 is called a translation
by a.
Translations are not linear transformations, since Ta (0) 6= 0. But translations are rigid
motions: ||Ta (x) − Ta (y)|| = ||x − a − (y − a)|| = ||x − y||.
1
Theorem 1: Let V be an n-dimensional i.p.s. over R. Then there is a distance-preserving
isomorphism φ : V → Rn , where the scalar product in Rn is given by the dot product.
Proof: Choose an orthonormal basis {e1 , . . . , en } for V . Then for x, y ∈ V , we have
n
X n
X
< x, y > = < < x, ei > ei , < y, ej > ej >
i=1 j=1
n X
X n
= xi y j < e i , e j >
i=1 j=1
n X
X n
= xi yj δij
i=1 j=1
n
X
= xi y i ,
i=1
where xi , yi are the coordinates of x, y in the given orthonormal basis. Now define
n
X
φ(x) = xi Ei ∈ Rn ,
i=1
n
where {Ei : 1 ≤ i ≤ n}
Pnis the standard orthonormal basis in R . Then clearly (a) φ is linear,
and (b) φ(x)•φ(y) = i=1 xi yi = < x, y > . It follows from this that ||φ(x)−φ(y)|| = ||s−y||
(why?) so the proof is complete
Definition: A vector space isomorphism which preserves the distances between vectors is
called an isometry.
Note:
• The isometry is far from unique. We can use the same construction with any o.n. basis
in V and any o.n. basis in Rn .
• As a consequence of the theorem, any time we want to demonstrate something about
a finite-dimensional real inner product space, it suffices to demonstrate it on En , since
the result can be transferred to V via the isometry φ−1 . So we don’t give up any
generality by working in En .
• The theorem as stated is also true for finite-dimensional complex inner product spaces.
There are isometries between them and Cn with the standard scalar product < z, w >=
w∗ z.
2
Theorem: Suppose that G is a linear transformation on En which preserves the scalar
product. Then G is orthogonal.
Proof: We’re given that Gx•Gy = x•y, ∀x, y. Then x•G∗ Gy = x•y. Moving everything
to the left hand side, we get
x•G∗ Gy − x•y = x• G∗ Gy − y = 0.
For any given y, this equality holds for all x ∈ En . This means that for any given y, we
must have G∗ Gy − y = 0, or G∗ Gy = y for any y. And this means that G∗G = I, so G is
orthogonal
Theorem: Any rigid motion F in En may be written in the form F (x) = (Ta ◦ U )(x), where
U is orthogonal and Ta is a translation.
Proof: Given F , we define G by G(x) = F (x) − F (0). Then since ||G(x) − G(y)|| =
||F (x) − F (y)|| = ||x − y||, G is a rigid motion that satisfies G(0) = 0. We’ll first show
that G preserves the scalar product, and then that G is linear (which by the theorem above
means that G is orthogonal:
(a) Since G is a rigid motion, G(x) − G(y) • G(x) − G(y) = (x − y)•(x − y). On the left
hand side of this, we have
G(x) − G(y) • G(x) − G(y) = ||G(x)||2 + ||G(y)||2 − 2G(x)•G(y)
= ||x||2 + ||y||2 − 2G(x)•G(y) (because G is a rigid motion)
On the right hand side,
||x − y||2 = ||x||2 + ||y||2 − 2x•y,
from which we conclude that G(x)•G(y) = x•y, so G preserves the scalar product.
(b) We now show that G is linear:
||G(x + cy) − G(x) − cG(y)||2 = ||[G(x + cy) − G(x)] − cG(y)||2
= ||G(x + cy) − G(x)||2 + c2 ||G(y)||2 − 2c G(x + cy) − G(x) •G(y)
= ||(x + cy) − x||2 + c2 ||y||2 − 2cG(x + cy)•G(y) + 2cG(x)•G(y)
= 2c2 ||y||2 − 2c(x + cy)•y + 2cx•y
= 2c2 ||y||2 − 2cx•y − 2c2 ||y||2 + 2cx•y
= 0,
where we’ve used the facts that G is a rigid motion and that G preserves the scalar product.
Comment: This last theorem gives another way to think about Euclidean geometry. If you
examine your high school geometry text carefully, you might conclude (depending on the
text of course), that the important term congruent is not really well-defined. For instance,
Wikipedia (not exactly the basic reference for mathematics) says that 2 figures are congruent
“if they have the same shape and size”, whatever that means. What they should have said
is that 2 figures A and B are congruent if there’s a rigid motion F such that F (A) = B.
